The gravitational field strength is approximately equal to 10 N.
<u>Explanation:</u>
Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.
So the gravitational field strength will be equal to the gravitational force acting on the object.
The formula for gravitational field strength is
Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.
Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.
Then the g = F
And 
Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.
So, the gravitational field strength is approximately equal to 10 N.
Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
Answer:
A) OA, AB, BC
B) 25m/s^2
C) see explanation
D) 25
E) Rest
Explanation:
From the Velocity time graph shown:
The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.
Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.
-ve slope = BC
B) Acceleration of body in path OA.
Acceleration = change in Velocity / time
Acceleration = (150 - 0) / 6
Acceleration = 150/6 = 25m/s^2
C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).
D) Length of BC
BC corresponds to the distance moved, that velocity / time
Velocity = 150 ; time = 6
Therefore Distance (BC) = 150/6 = 25
E.) Velocity =0 ; Hence body is at rest
Answer:
Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works
answer :
If the electrons falls through a PD of 150mV the electron will gain energy of 150MeV
Explanation:
when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron ) the electron will gain energy ( eV ) which is is equivalent in value of the PD it accelerated through
hence if the electrons falls through a PD of 150mV the electron will gain energy of 150MeV
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